e.g. 1 1996 HKCE Human Biology I (1aiii: modified)

The diagram below shows a pedigree for the inheritance of a human genetic disease.

From part (i), the disease is recessive and from (ii), the disease is sex-linked. Deduce, with reasons, the genotype of individual 2.

Genotype of female is XX. 5 has the disease. She must have two X-chromosomes with the recessive alleles for the trait i.e. she is homozygous recessive for the trait.

One of her X-chromosomes with the recessive allele must have been inherited from individual 2.

Individual 2 is normal in phenotype. Therefore, she must have an X-chromosome with the dominant normal allele.

Therefore, individual 2 is heterozygous.

e.g. 2 2012 HKDSE PP IB (9bi)

The female (B) has a family history of colour blindness. The son (E) has colour blindness. The pedigree of this family is shown in the following diagram:

In humans, colour blindness is a sex-linked trait. Based on the above pedigree, deduce the genotype of the mother (B) with respect to colour vision.

Being a sex-linked trait, the allele for colour blindness is located on the X-chromosome.

As individual E has colour blindness, he must have an X-chromosome with the recessive allele for colour blindness.

This X-chromosome must have been inherited from his mother B.

Genotype of female is XX. As the mother is normal, she must have at least one X-chromosome with the dominant normal allele for normal colour vision.

Therefore, the mother is heterozygous.

e.g. 3 2013 HKDSE IB (4a)

Red-green colour blindness is an X-linked recessive trait in humans. Peter is red-green colour blind while his daughter, Mary, is normal. Deduce Mary’s genotype without using a genetic diagram.

As Mary’s father is red-green colour blind, he must have an X-chromosome bearing the recessive allele for red-green colour blindness.

This X-chromosome must have been inherited to Mary.

Genotype of female is XX. As Mary is normal, she must have at least one X-chromosome with the dominant normal allele for normal eye sight.

Therefore, Mary is heterozygous.

e.g. 4 2013 HKAL II (4c)

In humans, a genetic disease D is caused by an allele located on the X chromosome. A couple, who are both without disease D, have given birth to a son with the disease. Based on this information, deduce the genotype of the mother and whether the allele for disease D is dominant.

As the son has disease D, he must have an X-chromosome with the recessive allele for the disease.

This X-chromosome must have been inherited from his mother.

Genotype of female is XX. As the mother is normal, she must have at least one X-chromosome with the dominant normal allele.

Therefore, the mother is heterozygous.

Under heterozygous conditions, only the dominant allele will be expressed while the recessive allele will be masked. Since mother is normal, the allele for disease D is recessive.